🟣 Systems of Nonlinear Equations

• In Level 2, we learned about systems of linear equations (straight lines)
• In Level 3, we now look at systems where at least one equation is a curve (parabola, circle, hyperbola, etc.)

Example of a Nonlinear System

$$y = x^2 - 4$$

$$y = x + 2$$

One equation is quadratic; the other is linear. The solutions are the points where their graphs intersect.

Number of Solutions

The number of solutions depends on how the graphs intersect:

• ✅ One solution = one intersection point
• ✅ Two solutions = two intersection points
• ✅ Multiple solutions = several intersection points
• ❌ No solution = curves never meet

• One equation is quadratic (parabola)
• The other is linear (straight line)
• Solutions: intersection(s) of line & parabola
• Can have 0, 1, or 2 solutions

Example:

$y = x^2 - 4$ and $y = 2x$

• Both equations are quadratic
• Solutions: intersection(s) of two parabolas
• Can have 0, 1, 2, 3, or 4 solutions

Example:

$y = x^2 - 3$ and $y = -x^2 + 5$

• One equation is a circle
• The other is a line
• Solutions: intersection(s) where line touches/crosses the circle
• Can have 0 (line misses), 1 (tangent), or 2 (secant) solutions

Example:

$x^2 + y^2 = 25$ and $y = x + 3$

• Both equations are circles
• Can have 0 (separate), 1 (tangent), 2 (intersecting), or infinite (coincident) solutions

Example:

$x^2 + y^2 = 9$ and $(x-2)^2 + y^2 = 4$

Steps:

1. Solve one equation for a variable
2. Substitute into the other equation
3. Solve the resulting equation
4. Back-substitute to find remaining variable

Example

Solve: $y = x^2 - 4$ and $y = 2x + 1$

Step 1: Set equal (both = $y$):

$$x^2 - 4 = 2x + 1$$

Step 2: Move all terms to one side:

$$x^2 - 2x - 5 = 0$$

Step 3: Apply the quadratic formula:

$$x = \frac{2 \pm \sqrt{4 + 20}}{2} = \frac{2 \pm \sqrt{24}}{2} = 1 \pm \sqrt{6}$$

Step 4: Substitute each $x$ into $y = 2x + 1$:

For $x = 1 + \sqrt{6}$: $y = 2(1 + \sqrt{6}) + 1 = 3 + 2\sqrt{6}$

For $x = 1 - \sqrt{6}$: $y = 2(1 - \sqrt{6}) + 1 = 3 - 2\sqrt{6}$

Answer: $(1+\sqrt{6}, 3+2\sqrt{6})$ and $(1-\sqrt{6}, 3-2\sqrt{6})$

Best if equations have similar or manipulable terms

Example: Solve $x^2 + y^2 = 25$ and $y = x + 3$

Step 1: Substitute $y = x + 3$ into $x^2 + y^2 = 25$:

$$x^2 + (x+3)^2 = 25$$

Step 2: Expand:

$$x^2 + x^2 + 6x + 9 = 25$$

$$2x^2 + 6x - 16 = 0$$

$$x^2 + 3x - 8 = 0$$

Step 3: Use quadratic formula to find $x$, then find $y$ from $y = x + 3$

• Graph both equations on the same coordinate plane
• Identify intersection points visually
• Useful for estimating solutions or verifying algebraic results

1️⃣ Physics: Projectile Motion

Finding when two objects at different trajectories meet

Example: Ball's height $h(t) = -5t^2 + 20t + 30$

Find when $h(t) = 40$

2️⃣ Economics: Supply & Demand

Market equilibrium where supply and demand curves intersect

Both can be nonlinear functions of price

3️⃣ Engineering: Circuit Design

Finding operating points in electronic circuits

Parabolic reflectors and antenna design

4️⃣ Astronomy: Orbital Mechanics

Computing intersection of orbital paths (ellipses)

Predicting satellite or planetary positions

5️⃣ GPS & Navigation

Trilateration uses systems of circles (spheres in 3D)

Your position is found at the intersection of multiple circles